面对最小割之类的题目，完全木想法。。。
枚举+最大流。。复杂度很大了。。。居然很快的就过了。。
1 /* 2 ID: cuizhe 3 LANG: C++ 4 TASK: telecow 5 */ 6 #include <cstring> 7 #include <cstdio> 8 #include <queue> 9 #include <algorithm> 10 using namespace std; 11 #define INF 0xffffff 12 struct node 13 { 14 int u,v,w,next; 15 } edge[10001]; 16 int first[1001],flag[1001],dis[1001]; 17 int qu[2001],qv[2001]; 18 int t,n,m,ans; 19 int sv,ev; 20 void CL() 21 { 22 t = 0; 23 memset(first,-1,sizeof(first)); 24 } 25 void add(int u,int v,int w) 26 { 27 edge[t].u = u; 28 edge[t].v = v; 29 edge[t].w = w; 30 edge[t].next = first[u]; 31 first[u] = t ++; 32 33 edge[t].u = v; 34 edge[t].v = u; 35 edge[t].w = 0; 36 edge[t].next = first[v]; 37 first[v] = t ++; 38 } 39 void build() 40 { 41 int i; 42 CL(); 43 for(i = 0; i < m; i ++) 44 { 45 add(qu[i]+n,qv[i],INF); 46 add(qv[i]+n,qu[i],INF); 47 } 48 for(i = 1; i <= n; i ++) 49 { 50 if(!flag[i]) 51 { 52 if(i == sv) 53 add(i,i+n,INF); 54 else if(i == ev) 55 add(i,i+n,INF); 56 else 57 add(i,i+n,1); 58 } 59 } 60 } 61 int bfs() 62 { 63 int u,v,i; 64 memset(dis,-1,sizeof(dis)); 65 queue<int> que; 66 que.push(sv); 67 dis[sv] = 0; 68 while(!que.empty()) 69 { 70 u = que.front(); 71 que.pop(); 72 for(i = first[u]; i != -1; i = edge[i].next) 73 { 74 v = edge[i].v; 75 if(edge[i].w > 0&&dis[v] < 0) 76 { 77 dis[v] = dis[u] + 1; 78 que.push(v); 79 } 80 } 81 } 82 if(dis[ev] > 0) return 1; 83 else return 0; 84 } 85 int dfs(int u,int step) 86 { 87 int i,a = 0,v; 88 if (u == ev) return step; 89 for (i = first[u]; i != -1; i = edge[i].next) 90 { 91 v = edge[i].v; 92 if (edge[i].w > 0&& dis[v] == dis[u]+1&&(a = dfs(v,min(step,edge[i].w)))) 93 { 94 edge[i].w -= a; 95 edge[i^1].w += a; 96 return a; 97 } 98 } 99 return 0; 100 } 101 int dinic() 102 { 103 int res,ans = 0; 104 while(bfs()) 105 { 106 while(res = dfs(sv,INF)) 107 ans+= res ; 108 } 109 return ans; 110 } 111 void path(int x) 112 { 113 int i; 114 if(x == 0) return ; 115 for(i = 1; i <= n; i ++) 116 { 117 if(flag[i] == 1) continue; 118 if(i == sv||i == ev) continue; 119 flag[i] = 1; 120 build(); 121 if(dinic() == x-1) 122 { 123 if(x == ans) 124 printf("%d",i); 125 else 126 printf(" %d",i); 127 path(x-1); 128 return ; 129 } 130 else 131 flag[i] = 0; 132 } 133 return ; 134 } 135 int main() 136 { 137 int i; 138 freopen("telecow.in","r",stdin); 139 freopen("telecow.out","w",stdout); 140 scanf("%d%d%d%d",&n,&m,&sv,&ev); 141 for(i = 0; i < m; i ++) 142 { 143 scanf("%d%d",&qu[i],&qv[i]); 144 } 145 build(); 146 printf("%d ",ans = dinic()); 147 path(ans); 148 printf(" "); 149 return 0; 150 }